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Title:Range Addition II 598
Difficulty:Easy
原题leetcode地址: https://leetcode.com/problems/range-addition-ii/
1. 问题转化为球ops中两行一维数组中最小值,然后将最小值相乘
时间复杂度:O(n),一次一层for循环,循环最长为n。
空间复杂度:O(1),没有申请额外空间。
/**
* 问题转化为球ops中两行一维数组中最小值,然后将最小值相乘
* @param m
* @param n
* @param ops
* @return
*/
public static int maxCount(int m, int n, int[][] ops) {
if (ops == null || ops.length == 0) {
return m * n;
}
int num1Min = Integer.MAX_VALUE;
int num2Min = Integer.MAX_VALUE;
for (int[] op : ops) {
num1Min = Math.min(num1Min, op[0]);
num2Min = Math.min(num2Min, op[1]);
}
return num1Min * num2Min;
}